Chapter 0.5, Problem 43E

To determine

To find: The inverse $f^{-1}$ of a function and to verify $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=\left(f^{-1}\text{ο}f\right)\left(x\right)=x$ .

Answer to Problem 43E

The inverse of function $f\left(x\right)=\frac{100}{1+2^{-x}}$ is $f^{-1}\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)$ .

Explanation of Solution

Given data: The given function is $f\left(x\right)=\frac{100}{1+2^{-x}}$ .

Method/Formula used:

Substitute y for $f\left(x\right)$ and find x in terms of y , write x as $f^{-1}\left(y\right)$ , then replace y by x throughout. The composite function $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=f\left[f^{-1}\left(x\right)\right]$ and $\left(f^{\text{-1}}\text{ο}f\right)\left(x\right)=f^{-1}\left[f\left(x\right)\right]$

Calculation:

The given function is $f\left(x\right)=\frac{100}{1+2^{-x}}$ .

Substitute y for $f\left(x\right)$ and solve for x as:

  $\begin{array}{c}y=\frac{100}{1+2^{-x}}\\ 1+2^{-x}=\frac{100}{y}\\ 2^{-x}=\frac{100}{y}-1\\ \frac{1}{2^x}=\frac{100-y}{y}\\ 2^x=\frac{y}{100-y}\end{array}$

Taking log of both sides.

  $\begin{array}{c}\ln \left(2^x\right)=\ln \left(\frac{y}{100-y}\right)\\ x\ln 2=\ln \left(\frac{y}{100-y}\right)\\ x=\frac{1}{\ln 2}\ln \left(\frac{y}{100-y}\right)\end{array}$

Replace, x by $f^{-1}\left(y\right)$ ,

  $f^{-1}\left(y\right)=\frac{1}{\ln 2}\ln \left(\frac{y}{100-y}\right)$

Change variable − replace y by customary variable x or another,

  $f^{-1}\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)$

Thus, inverse of the function $f\left(x\right)=\frac{100}{1+2^{-x}}$ is $f^{-1}\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)$ .

Now,

  $\left.\begin{array}{c}\left(f\text{ο}{f}^{-1}\right)\left(x\right)=f\left[f^{-1}\left(x\right)\right]\leftarrow \text{}\text{use}\text{definition}{f}^{-1}\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)\\ =f\left(\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)\right)\leftarrow \text{}\text{use}\text{definition}f\left(x\right)=\frac{100}{1+2^{-x}}\\ =\frac{100}{1+2^{-\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)}}\end{array}\right\}$

Now, solve $\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)$ . Let $\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)=p$ then

  $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=\frac{100}{1+2^{-p}}$ … (1)

Further,

  $\begin{array}{c}\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)=p\\ \ln \left(\frac{x}{100-x}\right)=p\ln 2\\ \ln \left(\frac{x}{100-x}\right)=\ln 2^p\end{array}$

Further,

  $\begin{array}{l}\ln \left(\frac{x}{100-x}\right)=\ln 2^p\\ \left(\frac{x}{100-x}\right)=2^p\\ \left(\frac{100-x}{x}\right)=2^{-p}\end{array}$

Substitute $\left(\frac{100-x}{x}\right)$ for $2^{-p}$ in result (1),

  $\begin{array}{c}\left(f\text{ο}{f}^{-1}\right)\left(x\right)=\frac{100}{1+\left(\frac{100-x}{x}\right)}\\ =\frac{100x}{x+100-x}\\ =x\end{array}$

Thus, $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=x$ verified.

Also,

  $\begin{array}{c}\left(f^{\text{-1}}\text{ο}f\right)\left(x\right)=f^{-1}\left[f\left(x\right)\right]\leftarrow \text{use}\text{definition}f\left(x\right)=\frac{100}{1+2^{-x}}\\ =f^{-1}\left[\frac{100}{1+2^{-x}}\right]\leftarrow \text{use}\text{definition}{f}^{-1}\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{x}{100-x}\right)\\ =\frac{1}{\ln 2}\ln \left(\frac{\frac{100}{1+2^{-x}}}{100-\frac{100}{1+2^{-x}}}\right)\\ =\frac{1}{\ln 2}\ln \left(\frac{\frac{100}{1+2^{-x}}}{\frac{100+100\cdot 2^{-x}-100}{1+2^{-x}}}\right)\end{array}$

Solving further,

  $\begin{array}{c}\left(f^{\text{-1}}\text{ο}f\right)\left(x\right)=\frac{1}{\ln 2}\ln \left(\frac{100}{100\cdot 2^{-x}}\right)\\ =\frac{1}{\ln 2}\ln \left(2^x\right)\\ =\frac{1}{\ln 2}\left(x\right)\ln 2\\ =x\end{array}$

Thus, the result $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=x=\left(f^{-1}\text{ο}f\right)\left(x\right)$ is verified.